Answer:
the work done by the 30N force is 4156.92 J.
For this problem, they don´t ask you to determine the work of the total force applied in the block. They only want the work done for the force of 30N, with an angle of 30º respectively of the displacement and a traveled distance of 160m. So:
W=F·s·cos(α)=30N·160m·cos(30º)=4156.92J
Answer:
it should be right it's from go.ogle hm!!!
Explanation:
Anterior or ventral - front (example, the kneecap is located on the anterior side of the leg). Posterior or dorsal - back (example, the shoulder blades are located on the posterior side of the body). Medial - toward the midline of the body (example, the middle toe is located at the medial side of the foot).
Current would flow between them and they would receive a terrible shock.
Answer:
λ = 2042 nm
Explanation:
given data
screen distance d = 11 m
spot s = 4.5 cm = 4.5 ×
m
separation L = 0.5 mm = 0.5 ×
m
to find out
what is λ
solution
we will find first angle between first max and central bright
that is tan θ = s/d
tan θ = 4.5 × / 11
θ = 0.234
and we know diffraction grating for max
L sinθ = mλ
here we know m = 1 so put all value and find λ
L sinθ = mλ
0.5 × sin(0.234) = 1 λ
λ = 2042.02 ×
m
λ = 2042 nm
Model I'm guessing. Coz that's using an object to explain
